Prove that there exists a semi-orthogonal $U$ such that $U^TAU=B$, where $A$ and $B$ are positive-definite...
0
$begingroup$
Let there be a semi-orthogonal matrix $U in mathbb{R}^{mtimes n}$ such that $U^TU=I_n$ if $m > n$
If $A in mathbb{R}^{mtimes m}$ and $B in mathbb{R}^{ntimes n}$ are positive-definite symmetric matrices,
How may I prove that there exists a $U$ such that
$$U^TAU=B quad ?$$Shall we call $U$ a change-of-basis matrix? i.e. changing the bases from dimension $n$ to $m$? OR Is there a better name to describe the transformation done by $U$?
Thanks in advance.
functional-analysis symmetric-groups orthogonality change-of-basis
asked Dec 31 '18 at 21:17
KayKay
698
$endgroup$
|
show 4 more comments
0
$begingroup$
Let there be a semi-orthogonal matrix $U in mathbb{R}^{mtimes n}$ such that $U^TU=I_n$ if $m > n$
If $A in mathbb{R}^{mtimes m}$ and $B in mathbb{R}^{ntimes n}$ are positive-definite symmetric matrices,
How may I prove that there exists a $U$ such that
$$U^TAU=B quad ?$$Shall we call $U$ a change-of-basis matrix? i.e. changing the bases from dimension $n$ to $m$? OR Is there a better name to describe the transformation done by $U$?
Thanks in advance.
functional-analysis symmetric-groups orthogonality change-of-basis
asked Dec 31 '18 at 21:17
KayKay
698
$endgroup$
2
$begingroup$
You have not explained the problem clearly. My understanding of the problem is this: we are given an $m times m$ symmetric matrix $A$ and an $n times n$ symmetric matrix $B$ (with $m > n$); we would like to prove that there exists a semi-orthogonal matrix $U$ such that $U^TAU = B$. Is that correct?
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:23
$begingroup$
As far as the terminology goes: $B$ is sometimes called a compression of $A$, but I don't know if there's any common term for the matrix $U$.
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:25
$begingroup$
this comes down to Sylvester's Law of Inertia. My first impression is that this is always possible if the number of positive eigenvalues of $A$ is at least as large as the count for $B,$ while the number of negative eigenvalues for $A$ is at least as large as that for $B.$ I can see how to make up for any zero eigenvalues of $B$ out of what remains. However, for example if $A$ is positive definite and $B$ is indefinite or negative, it cannot be done. Meanwhile, you should identify the exact book or set of notes you are using.
$endgroup$
– Will Jagy
Dec 31 '18 at 21:30
$begingroup$
Or, more trivially, take $A=0$ to see it's false in general.
$endgroup$
– Berci
Dec 31 '18 at 21:59
$begingroup$
@Omnomnomnom. Yes, that's right. I have edited the post accordingly. Thanks.
$endgroup$
– Kay
Dec 31 '18 at 22:27
|
show 4 more comments
0
0
0
$begingroup$
Let there be a semi-orthogonal matrix $U in mathbb{R}^{mtimes n}$ such that $U^TU=I_n$ if $m > n$
If $A in mathbb{R}^{mtimes m}$ and $B in mathbb{R}^{ntimes n}$ are positive-definite symmetric matrices,
How may I prove that there exists a $U$ such that
$$U^TAU=B quad ?$$Shall we call $U$ a change-of-basis matrix? i.e. changing the bases from dimension $n$ to $m$? OR Is there a better name to describe the transformation done by $U$?
Thanks in advance.
functional-analysis symmetric-groups orthogonality change-of-basis
asked Dec 31 '18 at 21:17
KayKay
698
$endgroup$
Let there be a semi-orthogonal matrix $U in mathbb{R}^{mtimes n}$ such that $U^TU=I_n$ if $m > n$
If $A in mathbb{R}^{mtimes m}$ and $B in mathbb{R}^{ntimes n}$ are positive-definite symmetric matrices,
How may I prove that there exists a $U$ such that
$$U^TAU=B quad ?$$Shall we call $U$ a change-of-basis matrix? i.e. changing the bases from dimension $n$ to $m$? OR Is there a better name to describe the transformation done by $U$?
Thanks in advance.
functional-analysis symmetric-groups orthogonality change-of-basis
functional-analysis symmetric-groups orthogonality change-of-basis
asked Dec 31 '18 at 21:17
KayKay
698
asked Dec 31 '18 at 21:17
KayKay
698
edited Dec 31 '18 at 22:51
Kay
asked Dec 31 '18 at 21:17
KayKay
698
asked Dec 31 '18 at 21:17
KayKay
698
asked Dec 31 '18 at 21:17
KayKay
698
698
2
$begingroup$
You have not explained the problem clearly. My understanding of the problem is this: we are given an $m times m$ symmetric matrix $A$ and an $n times n$ symmetric matrix $B$ (with $m > n$); we would like to prove that there exists a semi-orthogonal matrix $U$ such that $U^TAU = B$. Is that correct?
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:23
$begingroup$
As far as the terminology goes: $B$ is sometimes called a compression of $A$, but I don't know if there's any common term for the matrix $U$.
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:25
$begingroup$
this comes down to Sylvester's Law of Inertia. My first impression is that this is always possible if the number of positive eigenvalues of $A$ is at least as large as the count for $B,$ while the number of negative eigenvalues for $A$ is at least as large as that for $B.$ I can see how to make up for any zero eigenvalues of $B$ out of what remains. However, for example if $A$ is positive definite and $B$ is indefinite or negative, it cannot be done. Meanwhile, you should identify the exact book or set of notes you are using.
$endgroup$
– Will Jagy
Dec 31 '18 at 21:30
$begingroup$
Or, more trivially, take $A=0$ to see it's false in general.
$endgroup$
– Berci
Dec 31 '18 at 21:59
$begingroup$
@Omnomnomnom. Yes, that's right. I have edited the post accordingly. Thanks.
$endgroup$
– Kay
Dec 31 '18 at 22:27
|
show 4 more comments
2
$begingroup$
You have not explained the problem clearly. My understanding of the problem is this: we are given an $m times m$ symmetric matrix $A$ and an $n times n$ symmetric matrix $B$ (with $m > n$); we would like to prove that there exists a semi-orthogonal matrix $U$ such that $U^TAU = B$. Is that correct?
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:23
$begingroup$
As far as the terminology goes: $B$ is sometimes called a compression of $A$, but I don't know if there's any common term for the matrix $U$.
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:25
$begingroup$
this comes down to Sylvester's Law of Inertia. My first impression is that this is always possible if the number of positive eigenvalues of $A$ is at least as large as the count for $B,$ while the number of negative eigenvalues for $A$ is at least as large as that for $B.$ I can see how to make up for any zero eigenvalues of $B$ out of what remains. However, for example if $A$ is positive definite and $B$ is indefinite or negative, it cannot be done. Meanwhile, you should identify the exact book or set of notes you are using.
$endgroup$
– Will Jagy
Dec 31 '18 at 21:30
$begingroup$
Or, more trivially, take $A=0$ to see it's false in general.
$endgroup$
– Berci
Dec 31 '18 at 21:59
$begingroup$
@Omnomnomnom. Yes, that's right. I have edited the post accordingly. Thanks.
$endgroup$
– Kay
Dec 31 '18 at 22:27
2
2
$begingroup$
You have not explained the problem clearly. My understanding of the problem is this: we are given an $m times m$ symmetric matrix $A$ and an $n times n$ symmetric matrix $B$ (with $m > n$); we would like to prove that there exists a semi-orthogonal matrix $U$ such that $U^TAU = B$. Is that correct?
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:23
$begingroup$
You have not explained the problem clearly. My understanding of the problem is this: we are given an $m times m$ symmetric matrix $A$ and an $n times n$ symmetric matrix $B$ (with $m > n$); we would like to prove that there exists a semi-orthogonal matrix $U$ such that $U^TAU = B$. Is that correct?
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:23
$begingroup$
As far as the terminology goes: $B$ is sometimes called a compression of $A$, but I don't know if there's any common term for the matrix $U$.
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:25
$begingroup$
As far as the terminology goes: $B$ is sometimes called a compression of $A$, but I don't know if there's any common term for the matrix $U$.
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:25
$begingroup$
this comes down to Sylvester's Law of Inertia. My first impression is that this is always possible if the number of positive eigenvalues of $A$ is at least as large as the count for $B,$ while the number of negative eigenvalues for $A$ is at least as large as that for $B.$ I can see how to make up for any zero eigenvalues of $B$ out of what remains. However, for example if $A$ is positive definite and $B$ is indefinite or negative, it cannot be done. Meanwhile, you should identify the exact book or set of notes you are using.
$endgroup$
– Will Jagy
Dec 31 '18 at 21:30
$begingroup$
this comes down to Sylvester's Law of Inertia. My first impression is that this is always possible if the number of positive eigenvalues of $A$ is at least as large as the count for $B,$ while the number of negative eigenvalues for $A$ is at least as large as that for $B.$ I can see how to make up for any zero eigenvalues of $B$ out of what remains. However, for example if $A$ is positive definite and $B$ is indefinite or negative, it cannot be done. Meanwhile, you should identify the exact book or set of notes you are using.
$endgroup$
– Will Jagy
Dec 31 '18 at 21:30
$begingroup$
Or, more trivially, take $A=0$ to see it's false in general.
$endgroup$
– Berci
Dec 31 '18 at 21:59
$begingroup$
Or, more trivially, take $A=0$ to see it's false in general.
$endgroup$
– Berci
Dec 31 '18 at 21:59
$begingroup$
@Omnomnomnom. Yes, that's right. I have edited the post accordingly. Thanks.
$endgroup$
– Kay
Dec 31 '18 at 22:27
$begingroup$
@Omnomnomnom. Yes, that's right. I have edited the post accordingly. Thanks.
$endgroup$
– Kay
Dec 31 '18 at 22:27
|
show 4 more comments
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2
$begingroup$
You have not explained the problem clearly. My understanding of the problem is this: we are given an $m times m$ symmetric matrix $A$ and an $n times n$ symmetric matrix $B$ (with $m > n$); we would like to prove that there exists a semi-orthogonal matrix $U$ such that $U^TAU = B$. Is that correct?
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:23
$begingroup$
As far as the terminology goes: $B$ is sometimes called a compression of $A$, but I don't know if there's any common term for the matrix $U$.
$endgroup$
– Omnomnomnom
Dec 31 '18 at 21:25
$begingroup$
this comes down to Sylvester's Law of Inertia. My first impression is that this is always possible if the number of positive eigenvalues of $A$ is at least as large as the count for $B,$ while the number of negative eigenvalues for $A$ is at least as large as that for $B.$ I can see how to make up for any zero eigenvalues of $B$ out of what remains. However, for example if $A$ is positive definite and $B$ is indefinite or negative, it cannot be done. Meanwhile, you should identify the exact book or set of notes you are using.
$endgroup$
– Will Jagy
Dec 31 '18 at 21:30
$begingroup$
Or, more trivially, take $A=0$ to see it's false in general.
$endgroup$
– Berci
Dec 31 '18 at 21:59
$begingroup$
@Omnomnomnom. Yes, that's right. I have edited the post accordingly. Thanks.
$endgroup$
– Kay
Dec 31 '18 at 22:27